Day 11, Mat227

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Announcements:
- Your third assignment is returned: graded 4, 5 and 16.
  - #4: lots to do there! Make sure that you attempt all parts.
    One of the most important aspects of this problem is...
    symmetry
    Exploit symmetry any time that you can.
    One strategy for doing this problem is to model the function given (and it seems that I alone -- sniff -- decided to proceed that way). The given function looks a lot like a cosine, with period of 12, and amplitude 3.
    By the way, once we have this model, we can evaluate the anti-derivative using substitution!
  - #5: straightforward confirmation of the FTC. You were to sketch first (amazing how many didn't). Make sure that your parabolas look parabolic!
    Austin or Joey might show us some nice integrals and derivatives.
  - #16: Lexie and Andrew H. made good use of substitution here. They used the chain rule, but from a little different direction than I did.
    Your answer shouldn't have any ${\theta}$ in it. ${\theta}$ is a dummy variable of integration....
- Your assignment over 4.4 is due Monday.
- You have a new assignment over 4.5, due Monday, 2/16 (after our exam).
I'll start with any questions over chapter 4 through 4.4. That is the material your first exam will cover.
Then we'll return to Section 4.5: The substitution rule (for solving selected integrals analytically)
- The substitution rule is the chain rule backwards. Every differentiation rule is written backwards to create an integration rule. Substitution is the flip side of the chain rule.
  - Q. What is the derivative of ${\huge \sin({x^3})$ ?
  - A: ${\huge \cos(x^3)3x^2}$
  - Hence,
    
    $\int{\cos(x^3)3x^2dx}=\sin(x^3)+C$
  More generally:
  - Q. What is the derivative of ${\huge f{(g(x))}$ ?
  - A: ${\huge f'(g(x))g'(x)}$
  - Hence,
    
    $\int{f'(g(x))g'({x})dx}=f(g(x))+C$
  That's the general idea. So, in terms of a definite integral, the rule is that
  $\int_{a}^{b}h'(g(x))g'(x)dx = h(g(b))-h(g(a))$ and, even better in some minds, make a substitution of variable: ${\int_{a}^{b}h'(g(x))g'(x)dx=\int_{g(a)}^{g(b)}h'(u)du=h(g(b))-h(g(a))}$
  We can always do the change of variables
  
  ${\int_{a}^{b}f(g(x))g'(x)dx=\int_{g(a)}^{g(b)}f(u)du}$
  and hope that the integral on the right is easier to solve (certainly less cluttered). Notice especially the change in the limits on the integral.
  Writing it in this last way may be mysterious, because of the change of variable to u (and the change in the limits); but it's the disappearance of g'(x) that's really curious. It falls right out of the change of variables, however:
  ${u=g(x)\Longrightarrow\frac{du}{dx}=g'(x)}$ Solving for du, we get $du=g'(x)dx$ and hence that piece of the integrand disappears.
- Examples:
  - #2, p. 335
  - #5
  - #8
  - #19
  - #38
  - #57

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